Here is the question I want to answer:
Let $R = \mathbb Z[\sqrt{-5}]$
Show that $I = (2, 1 + \sqrt{-5})$ is not a free $R-$module.
Hint: It helps to view $R$ as $\mathbb Z[x]/(x^2 + 5).$
My questions are::
1- How can I use the hint?
I have found here How to show that an $\mathbb{Z}[\sqrt{-5}]$-module is not free? how to show that it is not free as it says in the answer in that link the following:
It is generated by two elements so if it were free, it would be free of rank $1$ or $2$.
If it were free of rank $1$, it would be generated by a single element: you can then try to prove that no element of $R$ divides both $2$ and $1+\sqrt{-5}$.
If it were free of rank $2$, then $2$ and $1+\sqrt{-5}$ would have to be free generators (can you see why ?). You can then try to prove that this is not the case, that is, that $2$ and $1+\sqrt{-5}$ are not linearly independent.
2- for showing this "If it were free of rank $1$, it would be generated by a single element: you can then try to prove that no element of $R$ divides both $2$ and $1+\sqrt{-5}$.", should I use the norm function?
Could someone push me in the right direction to answer those questions please?
Hint: Any two elements of $R$ are linearly dependent. (There is an obvious relation. Can you see this?)
From this, can you conclude that an ideal is free only if it is principal?
Edit: To see the first part, for distinct $a, b \in R$, note that $$(-b)\cdot a + a \cdot b = 0$$ and not both $a$ and $-b$ are $0$.
The second part now follows because any $R$-basis of $I \subset R$ must now either be $\varnothing$ or a singleton.